Question

A researcher is interested in looking at a new type of insulin that can treat diabetics with high fasting glucose levels. Suppose the researcher wants to take a sample of participants from the population of participants in the study in order to assess the progress by calculating their average resting glucose level (mg/dL). The researcher wants to be 95% confident using a population standard deviation of 27 mg/dL and hopes to have a margin of error at plus or minus 5 mg/dL. Calculate the sample size the researcher should get in order to answer his research question with all of the given information.

Answer 1

Answer:

The sample size should be at least 79.

Step-by-step explanation:

Let Y be be the resting glocose level of a member of the population chosen randomly, then Y is a random variable with unkown mean λ and Standard deviation σ = 27.

Let X be the sample mean of a sample of lenght n. X has the same mean as Y and the standard deviation is $\sigma = \frac{27}{\sqrt{n}} .$

If n is reasonable high, the Central Limit Theorem states that X has distribution approximately Normal, with mean  λ and Standard deviation $sigma = \frac{27}{\sqrt{n}} .$

If we standarize X, we get a random variable W

$W = \frac{X-\lambda}{\frac{27}{\sqrt{n}}} \simeq N(0,1)$

The values of W are tabulated and can be found on the attached file. We want a 95% confidence interval, so we want Z such that

P(-Z < w < Z) = 0.95

Using the symmetry of the normal density function, we get that

P(W < Z) = 0.975

If we look at the table, we will find that Z = 1.96, therefore we have

$P(-1.64 < \frac{X-\lambda}{\frac{27}{\sqrt{n}}} < 1.64) = 0.95$

Equivalently,

$0.95 = P(-1.64*\frac{27}{\sqrt{n}} < X- \lambda < 1.64*\frac{27}{\sqrt{n}})$

Taking out the X and the sign, after reverting the inequalities, we obtain

$P(X -1.64*\frac{27}{\sqrt{n}} < \lambda < X +1.64*\frac{27}{\sqrt{n}}) = 0.95$

Thus, a confidence interval with 95% confidence is

$CI = [ X-1.64*\frac{27}{\sqrt{n}}, X+1.64*\frac{27}{\sqrt{n}}]$

The (absolute) margin of error of this interval is $1.64*\frac{27}{\sqrt{n}} ,$ we want that number to be at most 5, so we take n such that

$1.64*\frac{27}{\sqrt{n}} = 5 \, \rightarrow \frac{1}{\sqrt{n}} = \frac{5}{1.64*27} \, \rightarrow \sqrt{n} = 8.85 \, \rightarrow n = 78.42$

We take n = 79. I hope that works for you!

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